java.math.BigInteger.add(BigInteger val)用于计算两个BigInteger的算术和。此方法用于查找比Java最大数据类型double的范围大得多的大量范围的算术加法,而不会影响结果的精度。此方法对当前的BigInteger执行操作,调用该方法并将BigInteger作为参数传递。
用法:
public BigInteger add(BigInteger val)
参数:此方法接受参数val,该参数是要添加到此BigInteger的值。
返回值:此方法返回一个保存有sum(this + val)的BigInteger。
下面的程序用于说明BigInteger的add()方法。
示例1:
// Java program to demonstrate
// add() method of BigInteger
import java.math.BigInteger;
public class GFG {
public static void main(String[] args)
{
// BigInteger object to store result
BigInteger sum;
// For user input
// Use Scanner or BufferedReader
// Two objects of String created
// Holds the values to calculate the sum
String input1
= "5454564684456454684646454545";
String input2
= "4256456484464684864864864864";
// Convert the string input to BigInteger
BigInteger a
= new BigInteger(input1);
BigInteger b
= new BigInteger(input2);
// Using add() method
sum = a.add(b);
// Display the result in BigInteger
System.out.println("The sum of\n"
+ a + " \nand\n" + b + " "
+ "\nis\n" + sum + "\n");
}
}
输出:
The sum of
5454564684456454684646454545
and
4256456484464684864864864864
is
9711021168921139549511319409
从上面的示例中可以看到,即使在添加了非常大的长度的数字之后,数据也将被完全精确化。
示例2:
// Java program to demonstrate
// add() method of BigInteger
import java.math.BigInteger;
public class GFG {
public static void main(String[] args)
{
// BigInteger object to store result
BigInteger sum;
// double variable
// To store result as double
double d;
// For user input
// Use Scanner or BufferedReader
// Two objects of String
// Holds the values to sum
// The number's length is of 400 digits
// Which is out of range of double
String input1 = "012345678901234567"
+ "8901234567890123"
+ "4567890123456789"
+ "0123456789012345"
+ "6789012345678901"
+ "2345678901234567"
+ "8901234567890123"
+ "4567890123456789"
+ "0123456789012345"
+ "6789012345678901"
+ "2345678901234567"
+ "8901234567890123"
+ "4567890123456789"
+ "0123456789012345"
+ "6789012345678901"
+ "2345678901234567"
+ "8901234567890123"
+ "4554324324362432"
+ "7674637264783264"
+ "7832678463726478"
+ "3264736274673864"
+ "7364732463546354"
+ "6354632564532645"
+ "6325463546536453"
+ "6546325463546534"
+ "6325465345326456"
+ "4635463263453264"
+ "654632498739473";
String input2 = "0123456789012345"
+ "6789012345678901"
+ "2345678901234567"
+ "8901234567890123"
+ "4567890123456789"
+ "0123456789012345"
+ "6789012345678901"
+ "2345678901234567"
+ "8901234567890123"
+ "4567890123456789"
+ "0123456789012345"
+ "6789012345678901"
+ "2345678901234567"
+ "8901234567890123"
+ "4567890123456789"
+ "0123456789012345"
+ "6789012345678901"
+ "2345873271893718"
+ "2974897146378481"
+ "7489127847281478"
+ "2174871248721847"
+ "8748327463756475"
+ "6745781641263981"
+ "2839721897438974"
+ "3286574365764576"
+ "2376914689217817"
+ "4362473624721647"
+ "61247612748612746";
// convert the string input to BigInteger
BigInteger a
= new BigInteger(input1);
BigInteger b
= new BigInteger(input2);
// Using add() method
sum = a.add(b);
// Display the result in BigInteger
System.out.println("The sum of\n"
+ a + " \nand\n" + b + " "
+ "\nis\n" + sum);
// Using double to hold the result
d = Double.parseDouble(sum.toString());
// Display the result in double
System.out.println("Using double, Sum is "
+ d);
}
}
输出:
The sum of
1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234554324324362432767463726478326478326784637264783264736274673864736473246354635463546325645326456325463546536453654632546354653463254653453264564635463263453264654632498739473
and
1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234587327189371829748971463784817489127847281478217487124872184787483274637564756745781641263981283972189743897432865743657645762376914689217817436247362472164761247612748612746
is
2469135780246913578024691357802469135780246913578024691357802469135780246913578024691357802469135780246913578024691357802469135780246913578024691357802469135780246913578024691357802469135780246913578024691357802469135780246913578024691357802469135780246913578024691357802469141651513734262516435190263143967454631918743000751861146858652219747883919392209327966909307740297653290433886520376204000415840169342671082000882825735618025902245247352219
Using double, Sum is Infinity
从上面的输出中可以明显看出,对于较大的整数使用double并不是一个好选择。
参考: https://docs.oracle.com/en/java/javase/12/docs/api/java.base/java/math/BigInteger.html#add(java.math.BigInteger)
相关用法
- Java BigInteger gcd()用法及代码示例
- Java BigInteger subtract()用法及代码示例
- Java BigInteger sqrtAndRemainder()用法及代码示例
- Java BigInteger intValueExact()用法及代码示例
- Java BigInteger nextProbablePrime()用法及代码示例
- Java 8 BigInteger longValueExact()用法及代码示例
- Java BigInteger isProbablePrime()用法及代码示例
- Java 8 BigInteger shortValueExact()用法及代码示例
- Java BigInteger multiply()用法及代码示例
- Java 8 BigInteger divideAndRemainder()用法及代码示例
- Java 8 BigInteger byteValueExact()用法及代码示例
- Java BigInteger divide()用法及代码示例
- Java BigInteger not()用法及代码示例
- Java BigInteger mod()用法及代码示例
- Java BigInteger and()用法及代码示例
注:本文由纯净天空筛选整理自Rajnis09大神的英文原创作品 BigInteger add() Method in Java with Examples。非经特殊声明,原始代码版权归原作者所有,本译文未经允许或授权,请勿转载或复制。